A2 June 2019 Q4

EdexcelCurrent spec12 marksCentres of Mass

4. A flagpole, \(AB\), is 4 m long. The flagpole is modelled as a non-uniform rod so that, at a distance \(x\) metres from \(A\), the mass per unit length of the flagpole, \(m\ \text{kg m}^{-1}\), is given by \(m = 18 - 3x\).

(a) Show that the mass of the flagpole is 48 kg. (3)

Figure 3: flagpole AB hinged at A on a vertical wall, at 45 degrees to the wall; a cable from the wall above A meets the pole at right angles at its midpoint; a ball at B — Figure 3

The end \(A\) of the flagpole is fixed to a point on a vertical wall. A cable has one end attached to the midpoint of the flagpole and the other end attached to a point on the wall that is vertically above \(A\). The cable is perpendicular to the flagpole. The flagpole and the cable lie in the same vertical plane that is perpendicular to the wall. A small ball of mass 4 kg is attached to the flagpole at \(B\). The cable holds the flagpole and ball in equilibrium, with the flagpole at 45° to the wall, as shown in Figure 3.

The tension in the cable is \(T\) newtons.

The cable is modelled as a light inextensible string and the ball is modelled as a particle.

(b) Using the model, find the value of \(T\). (8)

Mark scheme (a)

Scheme	Marks	AO
Total mass \(= \displaystyle\int_0^4 (18 - 3x)\,\mathrm{d}x\)	M1	2.1
\(= \left[18x - \dfrac{3x^2}{2}\right]_0^4\)	A1	1.1b
\(= 18\times 4 - \dfrac{3\times 16}{2}\ (= 72 - 24) = 48\) (kg) *	A1*	1.1b
	(3)

Notes

M1: Use integration (usual rules) – do not need to see limits at this stage

A1: (M1 on epen) Correct integration and correct limits seen

A1*: Show sufficient working to justify given answer

Mark scheme (b)

Scheme	Marks	AO
Taking moments about the base: \(\displaystyle\int_0^4 x(18 - 3x)\,\mathrm{d}x\)	M1	3.4
\(= \left[9x^2 - x^3\right]_0^4\ (= 80)\)	A1	1.1b
\(\Rightarrow 48d = 80\)	M1	3.4
\(d = \dfrac{80}{48} = \dfrac{5}{3}\) (m)	A1	1.1b
Complete strategy	M1	3.1b
\(\mathrm{M}(A):\ 2T = 4\cos 45^\circ\times 4g + \dfrac{5}{3}\cos 45^\circ\times 48g\)	A1ft	1.1b
\(\left(= \dfrac{96g}{\sqrt{2}}\right)\)	A1ft	1.1b
\(T = 333\) or 330	A1	2.2a
	(8)

Notes

M1: Use the model to find the moment of the pole and the ball about \(A\) (usual rules for integration)

A1: Correct integration

M1: Use the model to complete the moments equation.
Require their 80 and 48 used correctly

A1: Any equivalent form

M1: Complete strategy to find the tension – e.g.locate c of m of the pole and use moments.

A1ft: Moments equation with at most one error. Follow their c of m provided not at centre

A1ft: Correct unsimplified moments equation for their c of m not at centre

A1: Accept \(24\sqrt{2}g\), 333 or 330 ISW

Mark scheme (c)

Scheme	Marks	AO
Any appropriate comment e.g. the ball has been modelled as a point mass – its centre could be further from \(A\)	B1	3.5b
	(1)
	(12 marks)

Notes

B1:

The mass of the cable has been ignored – unlikely to be negligible if it can hold a pole this long.
The flagpole will be subject to cross winds
The cable might be extensible
The pole might not be rigid

NOT: the wall might be rough / smooth

Ignore incorrect statements